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Specifically, I am trying to wrap my head around this scenario but can't do so solely by thinking about the weight of the fluid.

Imagine a perfectly still lake. If you dive into the lake, the further down you go, the more pressure you feel - presumably because of the weight of the water above you. Now, imagine that I replace some arbitrary chunk of water beneath the surface with a (rigid) solid object that is magically held in place. If I were to swim and place myself beneath this object, one would naively think that there is less weight of water above me (after all, we got rid of some chunk of fluid), but in fact I feel the same pressure on me regardless of whether the object is there or not. In essence, it's the height I am beneath the surface that matters, not the weight of the fluid above me. Why? What is generating this force if not the weight?

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    $\begingroup$ This has been asked before. If the fluid consisted of a set of vertical solid rods, then each one could ignore its neighbors. But, pressure in a fluid "goes around corners". Take a tube of toothpaste, cap off, and hold it pointed upward. Now squeeze if from the sides. What happens? Some comes out the top. How is that? You didn't squeeze it up. You squeezed it sideways. $\endgroup$ – Mike Dunlavey Oct 11 '16 at 21:30
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The concept you need to include in your thinking is related to Pascal's Principle. As applied to your question, it means the pressure from the water above a particular position under water does not just act downward. That pressure acts in all directions. So, in effect, pressure in a fluid "moves around corners". When you are under the hypothetical fixed object in your scenario, you feel the pressure from the water to the sides of you. That water feels pressure from the water above as well as to the sides (and bottom) of it. If more water is added, or you and your object are deeper, the greater weight of more water above the water to the sides of will result in greater pressure there which will be transmitted in all directions including toward you.

Another way to look at this is if the pressure were less under the object, the greater pressure on the sides would "push" more against this lower pressure region until the pressures were equal.

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  • $\begingroup$ I agree with you, but this implies that the usual argument that it pressure is the "weight of the water column above" is oversimplified, The example given by the OP is a counterexample. $\endgroup$ – user126422 Oct 11 '16 at 17:40
  • $\begingroup$ Not sure what the counter example is. Also, the "usual" argument is that the pressure is DUE to the weight of the water above. Weight is a force, pressure is force per unit area. I added a sentence to my explanation, does that address your concern? $\endgroup$ – bpedit Oct 11 '16 at 18:51
  • $\begingroup$ is certainly not the weight of the water just above you $\endgroup$ – user126422 Oct 11 '16 at 20:07
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All of the other answers are correct. However I shall re-express their answer in terms of "weight of fluid above you", which perhaps you will find more palatable.

For simplicity say the solid body submerged in the fluid is an upright cylinder of cross-sectional area $A$, and has weight $W_{cylinder}$. We want to calculate pressure at some point in the fluid beneath the cylinder. To do so we shall calculate weight of fluid+solid-cylinder column above that point, acting on horizontal area of magnitude $A$, and then divide total force acting on that area by magnitude of area. Key to the argument is that total force acting on area is not simply equal to weight of fluid +solid-cylinder column above that area.

Total weight of fluid+solid-cylinder column=$W_{cylinder}+W_{fluid}$

But since the cylinder is held stationary, then by virtue of Archimedes principle, we must exert an upward force on it equal to $W_{cylinder}-W_{displaced~fluid}$, where $W_{displaced~fluid}$ is simply the weight of fluid displaced by cylinder.

So net downward force on area $A$ is \begin{align} (W_{cylinder}+W_{fluid})-(W_{cylinder}-W_{displaced~fluid})=W_{fluid}+W_{displaced~fluid} \end{align} which is nothing but weight of water column in which the solid cylinder has been replaced by fluid cylinder. That is why, to calculate pressure, it does not matter whether there is a solid object hanging above a point inside the fluid, provided the whole system is in equilibrium.

P.S. I have assumed that cylinder is denser than fluid. I shall leave it to you to generalize the argument.

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Pressure is (directly dependent on) the weight of fluid above you. Weight takes into account the density of the fluid, gravity and volume of the fluid. The problem you are thinking about can be broken down into simple understandable bits. If water above you had a solid of the same (or lower density in case you have filled a cavity in the solid with some heavy stones) the effect would still be the same primarily because the system as a whole is in equilibrium and static. What that really means is that along the surface at a particular depth, the pressure will be the same irrespective of the fact that above some portion of the surface a different object is placed. The system will displace the excess fluid, and the solid (if needed) until the above condition is not achieved. Because if there was a pressure difference, then it would no longer be a static system but a dynamic or a transient situation. The amount of water displaced.

The pressure difference across the body (say a cube) will be (density of water)(side length of the cube)(gravity) which will be equal to the pressure exerted by the box (due to its weight) on that piece of the surface. The two pressures will be equal but will arise due to two equal forces acting in opposite directions, thus bringing about an equilibrium and also making the pressure at thag depth same as that due to water.

In case you didn't understand why there is pressure due to weight: consider a cylinder filled to the brim with a liquid. At the bottom, the net force applied by water will be the Maas of water times gravity. Thus the pressure will be the force divided by the area. Now, we know that the mass is volume of liquid times the volume which is area times height. Thus in the expression of pressure, the area term will cancel out leaving behind just pressure = (density)(height)(gravity). So, in a sense (and well always) pressure will be due to the weight/normal force at the surface.

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  • $\begingroup$ Pressure is DUE to the weight of the fluid above you. Weight is a force, pressure is force per unit area. $\endgroup$ – bpedit Oct 11 '16 at 19:15
  • $\begingroup$ Yeah, I guess that is what I said $\endgroup$ – Pranshu Malik Oct 12 '16 at 3:21
  • $\begingroup$ You might want clarify in your first sentence. The topic is pressure, when you lead off with "It", a reader assumes that's what you mean. $\endgroup$ – bpedit Oct 12 '16 at 4:05
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Pressure at some depth of fluid is exerted in all directions, up, down, sideways, and everything in between. This is because the molecules can't just squeeze out to the sides when the weight of fluid above pushes down, since the ones next to them are being pushed down too. So they all push against each other in all directions. That's why it doesn't matter if you're under the block.

A gas exerts its pressure on a surface (like you) by the action of gazillions of molecules bouncing off of you at any instant in their random motion. During the bounce, you exert a force on each molecule that changes its momentum, so it also exerts a force on you. The sum of all those little forces on an area at any instant divided by the area is pressure.

In a liquid, there's another effect. The repulsive electric fields between the molecules are compressed like springs. Although water is considered virtually incompressible, the compression is not zero, and a small reduction in distance occurs against a relatively large force. Molecules adjacent to a surface transfer that electrostatic force, and together exert a steady state force on it in addition to the collision forces. This explains the great pressures in the depths of the ocean, which can't be explained by greatly increased collision rates since it's cold down there and the density is isn't much greater.

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protected by Qmechanic Oct 12 '16 at 21:17

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